Sol. Clearly, dom (f) = R

Now,

f : R → R is invertible

⇒ f : R → R is bijection

⇒ f(x) is either strictly increasing or strictly decreasing on R

⇒ f ′ (x) > 0 or, f ′ (x) < 0 for all x ∈ R

⇒ 3x 2 + 2x (a + 2) + 3a > 0 or < 0 for all x ∈ R

But, 3x 2 + 2x (a + 2) + 3a cannot be less than zero for all x ∈ R. Because the curve y = 3x 2 + 2x (a + 2) + 3a is a parabola opening upward. Therefore, y > 0 for same x ∈ R. Consequently

3x 2 + 2x (a + 2) + 3a > 0 for all x ∈ R

⇒ 4(a + 2) 2 –36a < 0 [  disc. < 0]

⇒ a 2 + 4a + 4 –9a < 0

⇒ a 2 –5a + 4 < 0

⇒ (a –1) (a – 4) < 0

⇒ 1 < a < 4

⇒ a ∈ (1, 4)

Hence, f(x) is invertible, if a ∈ (1, 4)